Question Number 212001 by mnjuly1970 last updated on 26/Sep/24
$$ \\ $$$$\:\:\:\:{prove}\:\:{that}: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\underset{{k}\in\mathbb{Z}} {\sum}\:\frac{\:\left(−\mathrm{1}\right)^{{k}} }{\:{x}\:+\:{k}\pi}\:=\:\frac{\mathrm{1}}{{sin}\left({x}\right)}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$
Answered by Berbere last updated on 27/Sep/24
![Σ_(k∈Z) ∫_0 ^1 (−1)^k t^(x−1+kπ) dt=Σ_(k∈Z) (((−1)^k )/(x+kπ))=S(x) S(x)=S_1 (x)−S_1 (−x)+(1/x) S_1 (x)=Σ_(k≥0) (((−1)^k )/(x+kπ))=Σ_(k≥0) ∫_0 ^1 (−t^π )^k t^(x−1) dt=∫_0 ^1 (t^(x−1) /(1+t^π ))dt =∫_0 ^1 ((t^(x−1) (1−t^π ))/(1−t^(2π) ))dt=∫_0 ^1 ((z^((x−1)/(2π)) (1−z^(1/2) ))/(1−z)).z^((1/(2π))−1) (dz/((2π))) =(1/((2π)))∫_0 ^1 (z^((x/(2π))−1) −z^((x/(2π))−(1/2)) )(1−z)^(−1) dz Ψ(1+z)=−γ+∫_0 ^1 ((1−t^(z−1) )/(1−t))dt =(1/((2π)))[Ψ((x/(2π))+(1/2))−Ψ((x/(2π)))] S=(1/((2π) ))[Ψ((x/(2π))+(1/2))−Ψ((x/(2π)))−Ψ((1/2)−(x/(2π)))+Ψ(−(x/(2π)))] =(1/(2π))π[cot(π((x/(2π))+(1/2))−[Ψ(1+(x/(2π)))−((2π)/x)]+Ψ(−(x/(2π)))] =(1/(2π))[−((2π)/x)+πcot((x/2)+(π/2))+πcot(π+(x/2))] =−(1/x)+(1/2)[((sin((x/(2())))/(cos((x/2))))+((cos((x/2)))/(sin((x/2))))]=−(1/x)+(1/(2cos((x/2))sin((x/2)))) =−(1/x)+(1/(sin(x))) S=S1−S_1 (−x)+(1/x)+(1/(sin(x)))=(1/(sin(x)))](https://www.tinkutara.com/question/Q212014.png)
$$\underset{{k}\in\mathbb{Z}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {t}^{{x}−\mathrm{1}+{k}\pi} {dt}=\underset{{k}\in\mathbb{Z}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{x}+{k}\pi}={S}\left({x}\right) \\ $$$${S}\left({x}\right)={S}_{\mathrm{1}} \left({x}\right)−{S}_{\mathrm{1}} \left(−{x}\right)+\frac{\mathrm{1}}{{x}} \\ $$$${S}_{\mathrm{1}} \left({x}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{x}+{k}\pi}=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{t}^{\pi} \right)^{{k}} {t}^{{x}−\mathrm{1}} {dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{x}−\mathrm{1}} }{\mathrm{1}+{t}^{\pi} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{x}−\mathrm{1}} \left(\mathrm{1}−{t}^{\pi} \right)}{\mathrm{1}−{t}^{\mathrm{2}\pi} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{z}^{\frac{{x}−\mathrm{1}}{\mathrm{2}\pi}} \left(\mathrm{1}−{z}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)}{\mathrm{1}−{z}}.{z}^{\frac{\mathrm{1}}{\mathrm{2}\pi}−\mathrm{1}} \frac{{dz}}{\left(\mathrm{2}\pi\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}\pi\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \left({z}^{\frac{{x}}{\mathrm{2}\pi}−\mathrm{1}} −{z}^{\frac{{x}}{\mathrm{2}\pi}−\frac{\mathrm{1}}{\mathrm{2}}} \right)\left(\mathrm{1}−{z}\right)^{−\mathrm{1}} {dz} \\ $$$$\Psi\left(\mathrm{1}+{z}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{{z}−\mathrm{1}} }{\mathrm{1}−{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}\pi\right)}\left[\Psi\left(\frac{{x}}{\mathrm{2}\pi}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\frac{{x}}{\mathrm{2}\pi}\right)\right] \\ $$$${S}=\frac{\mathrm{1}}{\left(\mathrm{2}\pi\right)\:}\left[\Psi\left(\frac{{x}}{\mathrm{2}\pi}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\frac{{x}}{\mathrm{2}\pi}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}\pi}\right)+\Psi\left(−\frac{{x}}{\mathrm{2}\pi}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\pi\left[{cot}\left(\pi\left(\frac{{x}}{\mathrm{2}\pi}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\left[\Psi\left(\mathrm{1}+\frac{{x}}{\mathrm{2}\pi}\right)−\frac{\mathrm{2}\pi}{{x}}\right]+\Psi\left(−\frac{{x}}{\mathrm{2}\pi}\right)\right]\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left[−\frac{\mathrm{2}\pi}{{x}}+\pi{cot}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)+\pi{cot}\left(\pi+\frac{{x}}{\mathrm{2}}\right)\right] \\ $$$$=−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{sin}\left(\frac{{x}}{\mathrm{2}\left(\right.}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}+\frac{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\right]=−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\frac{{x}}{\mathrm{2}}\right){sin}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{sin}\left({x}\right)} \\ $$$${S}={S}\mathrm{1}−{S}_{\mathrm{1}} \left(−{x}\right)+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{sin}\left({x}\right)}=\frac{\mathrm{1}}{{sin}\left({x}\right)} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 27/Sep/24
$${thanks}\:{alot}\:{Sir}… \\ $$
Commented by Berbere last updated on 27/Sep/24
$${Withe}\:{Pleasur} \\ $$